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Homework Help: Pin Jointed truss analysis

  1. Mar 4, 2010 #1
    Hi all, been here several times and found good help for several questions, but this time I'm out of luck, so here I go.

    1. The problem statement, all variables and given/known data
    By taking a section through the pin-jointed frame as shown, and considering the quilibrium of the structure to either the left or right of the section, what is the force in members CD, CH and IH?

    Each member of the frame has a length of 2.0 m so the height of the frame is 2 × sin 60º = 1.732 m.
    http://img705.imageshack.us/img705/2225/pjfasection.gif [Broken]
    I'm trying to solve this using method of sections, and none of the examples I've found seem to help, though I feel I'm missing something obvious with regards to using sine and cosine to find the vertical values.

    So you remove the right hand side, and have CD and IH in tension, and CH in compression.

    http://img38.imageshack.us/img38/9983/pjt2.jpg [Broken]

    And this is where I'm getting lost, trying to resolve the forces vertically in order to find force CH, since that is the only unknown vertical force. There's the 15KN being applied upwards, and 15KN being applied downwards, but the value for CH can't be 0, because its in compression so it has to have a value.

    Now, from my lectures notes I have http://img220.imageshack.us/img220/9484/pjt3.jpg [Broken] but that doesn't help much since it is dealing with the unknown vertical force being directed downwards. But the other problem I'm seeing is that if I were to just use the same formula, for my example it would end up as 15 = Csin60 + 15

    And that's obviously wrong, since Csin60 isn't 0!

    I fear I'm getting rather confused, I used to think I understood maths pretty well until I met these trusses! Thanks in advance for any help!
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Mar 4, 2010 #2


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    Welcome to PF!

    Hi iviv ! Welcome to PF! :smile:
    But it is 0 … your equations tell you so!

    Hint: Imagine the two top forces were 20 on the left and 10 on the right (instead of 15 and 15).

    Then the left half would be top-heavy, wanting to turn clockwise, while the right half would want to turn anti-clockwise.

    Would that put CH in tension or compression? And DH? :wink:
  4. Mar 4, 2010 #3
    Thanks for the welcome, but oh dear, I think you've lost me there! You're saying that the equations are right, and it does equal zero, so the member isn't in compression or tension?

    Also, a related question I'm trying to get my head around. I think I'm doing it right, but I seem to have an answer which doesn't make much sense. I tried to illustrate the problem in paint, showing why I'm really a surveyor and not an architecht :D

    http://img408.imageshack.us/img408/1285/stuckt.jpg [Broken]

    AH is simple, it has to be 10, GH is also 0.

    AG was found because AGsin45 = 10 so FG = 14.1
    And AB = AGcos45 so AG = 9.97?

    Carrying on from there, FG = AGcos45 as well, so is 9.97 again.
    BG = AGsin45 = 9.97

    And that doesn't look like it makes sense, because to resolve vertically at B it would be:

    9.97+BFsin45 = 10 so BF = 0.0424?

    I'm still working through the question, but that seems like a rather silly answer, so while I'm here I thought I'd ask to see if there's something I'm obviously doing wrong.

    Thanks again for your time!
    Last edited by a moderator: May 4, 2017
  5. Mar 4, 2010 #4


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    Hi iviv! :smile:

    erm :redface: … isn't cos45º = sin45º ? :wink:
  6. Mar 4, 2010 #5
    Yeah it is, but that doesn't change any of the maths in those equations, does it? I mean, you use sin to resolve horizontally, and cos to resolve vertically?
  7. Mar 4, 2010 #6


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    well … it does change your 9.97 to 10, and your 0.0424 to … ? :wink:
  8. Mar 4, 2010 #7
    Oh god, its a rounding error, isn't it? XD

    Well, that makes things add up a whole lot better now! Thanks!
  9. Mar 4, 2010 #8
    Do you understand the term take moments eg take moments about C in your post#1 diagram?
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